A classical functional generalization of the first Barnes lemma

TL;DR

This paper presents a simplified proof of a contour integral formula for the Gauss hypergeometric function, which generalizes Barnes's first lemma and offers an alternative to existing formulas.

Contribution

It introduces a more straightforward proof and a new generalization of Barnes's first lemma for hypergeometric functions.

Findings

Provides a simpler proof of the contour integral formula.

Generalizes Barnes's first lemma for hypergeometric functions.

Offers an alternative to existing integral formulas.

Abstract

We give a brief account and a simpler proof of a contour integral formula for the Gauss hypergeometric function. Such formula is alternative to Barnes's integral formula and generalizes the first Barnes Lemma.

Equations

$$F(a,b;c;1)=\frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}.$$

$$F(a,b;c;z)=\sum _{k=0}^{\infty }\frac{(a{)}_k(b{)}_k}{k!(c{)}_k}{z}^k,$$

$$\Gamma (z)=\underset{0}{\overset{\infty }{\int }}{t}^{z-1}{e}^{-t}\mathrm{d}t,$$

$$\log \Gamma (z)=(s-\frac{1}{2})\log s-s+\frac{1}{2}\log (2\pi )+o(1),$$

$$\Gamma (z)\Gamma (1-z)=\frac{\pi z}{\sin \pi z},\Gamma (z+1)=z\Gamma (z).$$

$$\frac{\Gamma (a)\Gamma (b)}{\Gamma (c)}F(a,b;c;z)=\frac{1}{2\pi i}\underset{-i\infty }{\overset{i\infty }{\int }}\frac{\Gamma (a+s)\Gamma (b+s)\Gamma (-s)}{\Gamma (c+s)}(-z{)}^s\mathrm{d}s,$$

$$\frac{1}{2\pi i}\underset{-i\infty }{\overset{i\infty }{\int }}\Gamma (a+s)\Gamma (b+s)\Gamma (c-s)\Gamma (d-s)=\frac{\Gamma (a+c)\Gamma (a+d)\Gamma (b+c)\Gamma (b+d)}{\Gamma (a+b+c+d)},$$

$$\begin{gathered} F(a,b;c;z)=\frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}F(a,b;1+a+b-c;1-z) \\ +\frac{\Gamma (c)\Gamma (a+b-c)}{\Gamma (a)\Gamma (b)}(1-z{)}^{c-a-b}F(c-a,c-b;1+c-a-b;1-z) \end{gathered}$$

$$\frac{\Gamma (a)\Gamma (c-a)\Gamma (b)\Gamma (c-b)}{\Gamma (c)}F(a,b;c;1-z)=\frac{1}{2\pi i}\underset{-i\infty }{\overset{i\infty }{\int }}\Gamma (a+s)\Gamma (b+s)\Gamma (c-a-b-s)\Gamma (-s)z^s\mathrm{d}s,$$

$$(-1)^{n}\frac{{\rm d}^{n}}{{\rm d}z^{n}}F(a,b;c;1-z)\Big{|}_{z=1}=\frac{(a)_{n}(b)_{n}}{(c)_{n}}=\frac{\Gamma(a+n)}{\Gamma(a)}\frac{\Gamma(b+n)}{\Gamma(b)}\frac{\Gamma(c)}{\Gamma(c+n)}.$$

$$\frac{1}{2\pi i}\underset{-i\infty }{\overset{i\infty }{\int }}\Gamma (a+s)\Gamma (b+s)\Gamma (c-a-b-s)\Gamma (-s)s{z}^{s-1}\mathrm{d}s,$$

$$(-1)^{n}\frac{{\rm d}^{n}}{{\rm d}z^{n}}\frac{1}{2\pi i}\int\limits_{-i\infty}^{i\infty}\Gamma(a+s)\Gamma(b+s)\Gamma(c-a-b-s)\Gamma(-s)z^{s}{\rm d}s\Big{|}_{z=1}\\ =\frac{1}{2\pi i}\int\limits_{-i\infty}^{i\infty}\Gamma(a+s)\Gamma(b+s)\Gamma(c-a-b-s)\Gamma(n-s){\rm d}s\qquad(n=0,1,2,\dots).$$

$$\frac{\Gamma (c-a)\Gamma (c-b)\Gamma (a+n)\Gamma (b+n)}{\Gamma (c+n)},$$

$$\begin{gathered} \sum _{n=0}^{\infty }\frac{({\alpha }_1{)}_n({\alpha }_2{)}_n({\alpha }_3{)}_n}{n!({\beta }_1{)}_n({\beta }_2{)}_n}=\frac{\Gamma ({\beta }_1)}{\Gamma ({\alpha }_1)\Gamma ({\beta }_1-{\alpha }_1)\Gamma ({\alpha }_2)\Gamma ({\beta }_1-{\alpha }_2)} \\ \times \frac{1}{2\pi i}\underset{-i\infty }{\overset{i\infty }{\int }}\Gamma ({\alpha }_1+s)\Gamma ({\alpha }_2+s)\Gamma ({\beta }_1-{\alpha }_1-{\alpha }_2-s)\Gamma (-s)F({\alpha }_3,-s;{\beta }_2;1)\mathrm{d}s, \end{gathered}$$

$$H_n(\xi ,p;v)=\sum _{j=0}^n\frac{(-n{)}_j(n+1{)}_j(\xi {)}_j}{j{!}^2(p{)}_j}{v}^j(n=0,1,2,\dots ).$$

$$\begin{gathered} \Gamma (p-q)\Gamma (q)\Gamma (p-\xi )\Gamma (\xi )H_n(\xi ,p;v) \\ =\frac{\Gamma (p)}{2\pi i}\underset{\sigma -i\infty }{\overset{\sigma +i\infty }{\int }}\Gamma (s)\Gamma (q-s)\Gamma (\xi -s)\Gamma (p-q-\xi +s)H_n(s,q;v)\mathrm{d}s, \end{gathered}$$

$$\sum _{n=0}^{\infty }{t}^n{H}_n(\xi ,p;v)=\frac{1}{1-t}F(\xi ,1/2;p;-4vt(1-t{)}^{-2}).$$

Full text

1. Introduction

The Gauss hypergeometric function (denoted by $F(a,b;c)$ throughout the present paper) has been deeply studied, and several integral representations can be found in books dealing with special functions (see e.g. [7, Sections 8.3, 8.8]). An important integral was discovered by Barnes (see formula (3) below), who build an alternative theory of the function $F(a,b;c)$ based on such integral formula. One useful feature of formulas of the type (3) relies in the possibility of applying the saddle point method to obtain a precise asymptotic estimate of the function involved (see the monography [5]). Another interesting property of (3) is that it possesses a wide range of extensions to generalized hypergeometric series (see [6, Sections 4.6, 4.7]).

The countour integral formula proved in the present paper is not new (see [8, Section 14.53] and [4, formula (15.6.7)]). However, we believe it is worth the present short note, because our proof appears to be simpler than that in [8], and is independent of Barnes’s integral formula (3). We remark that formula (6) encompasses (and, in the present note, relies on) the first Barnes lemma (see (4) below), whose proof in [2] is very similar to the proof of Barnes’s integral formula (3). Therefore our contribution allows one to use the residue theorem in the proof of the first Barnes Lemma only. After that, one can prove the contour integral formula (6) as in the present paper, and finally combine the two results to prove (3), with an argument similar to [8, Section 14.53], without applying the residue theorem a second time, as in [8] . Also, our argument is very simple but apparently has been generally overlooked in this context, and may have further applications.

The first Barnes lemma is often considered as an integral analogue of the Gauss summation formula

[TABLE]

In addition, formula (6) can be seen as an integral analogue of the formula connecting the values of hypergeometric functions of $z$ and $1-z$ (see (5) below), and this is precisely the context where (6) is used in [8, Section 14.53]. Let us also point out two formulas close to (6): the first one, obtained in 1939 by S.O. Rice for his function $H_{n}(\xi,p;v)={}_{3}F_{2}(-n,n+1,\xi;1,p;v)$ (see [3, Vol I, p.193]), and the second one, usually used in the proof af the second Barnes lemma (see e.g. [2, p.43]. We mention these formulas at the end of the present paper.

2. The main result and a few similar formulas

We denote by $(\xi)_{n}$ the product $\xi(\xi+1)\cdots(\xi+n-1)$ for any complex number $\xi$ and for any $n=1,2,\dots$, and we put $(\xi)_{0}=1$. We say that $\xi$ is admissible if $\xi$ is not a negative integer nor [math].

The Gauss hypergeometric function $F(a,b;c;z)$ is defined over the unit disc $|z|<1$ in the complex plane by the series

[TABLE]

where $a$, $b$ and $c$ are complex numbers and $c$ is admissible. Note that the series $F(a,b;c;z)$ may terminate: this happens when $a$ or $b$ are not admissible. In this case the function (2) is a polynomial in $z$, and could be defined even if $c$ is not admissible, provided that $\min\{a,b\}\leq c$.

Let $\Gamma(z)$ be the Euler gamma function, defined in the complex half-plane ${\rm Re}\,z>0$ by

[TABLE]

and extended to a meromorphic function in the complex plane, with simple poles at $z=-n$ with residue $\frac{(-1)^{n}}{n}$ ($n=0,1,2,\dots$), for example by splitting the integration path $(0,\infty)$ in the union of $(0,1)$ and $(1,\infty)$. Two main properties of the function $\Gamma(z)$ are important in the following: the Stirling formula

[TABLE]

valid for $|\arg z|<\pi-\delta$ for any $\delta>0$, and the functional equations

[TABLE]

The Barnes integral representation (see e.g. [1, Theorem 2.4.1]) of the function (2) is given by

[TABLE]

valid under the conditions that $|z|<1$, $z\not=0$ and $|\arg(-z)|<\pi$, and that $a$, $b$ and $c$ are admissible. The path $L$ of integration is curved, if necessary, in such a way that separates the poles $s=-a-n$ and $s=-b-n$ ($n=0,1,2,\dots$) at the left of $L$ from the poles $s=0,1,2,\dots$ at the right of $L$. In the sequel, we denote by $F(a,b;c;z)$ the analytic function defined for $z\notin[1,\infty)$ either by the series (2), if $|z|<1$, or by the integral (3), if $z\notin[0,\infty)$.

The first Barnes lemma (see e.g. [1, Theorem 2.4.2]) states that

[TABLE]

provided that $a+c$, $a+d$, $b+c$ and $b+d$ are admissible.

Using (3) and (4) one can prove (see [8, Sect. 14.53]) that

[TABLE]

Using (4) we can prove an integral formula that encompasses (5), which is a generalization of (1). For this reason we named formula (6) below a functional generalization of the first Barnes lemma.

Theorem 2.1**.**

[8, Section 14.53]** Let $a$, $b$, $c$ and $z$ be complex numbers such that $z\notin(-\infty,0]$, and that $a$, $c-a$, $b$, $c-b$ and $c$ are admissible. Then

[TABLE]

where the integration path $L$ separates the poles $s=-a-n$ and $s=-b-n$ $(n=0,1,2,\dots)$ on the left of $L$ from the poles $s=n$ and $s=a+b-c+n$ $(n=0,1,2,\dots)$ on the right of $L$.

Proof.

Suppose that $|1-z|<1$. For any $n=0,1,2,\dots$ we have

[TABLE]

The integral at the right-hand side of (6) is an analytic function in the domain $|\arg z|<2\pi$ (see [5, Lemma 2.4]), which plainly contains the disc $|1-z|<1$. This implies that the derivative of the integral in (6) with respect to $z$ equals

[TABLE]

this being an integral of the same type as in (6), once it is noticed that $-s\Gamma(-s)=\Gamma(1-s)$, and after substituting the variable $s$ with $t$ by putting $s=1+t$, and then renaming $t$ with $s$. We thus have

[TABLE]

By (4) the last integral equals

[TABLE]

therefore (6) is proved for $|1-z|<1$, because all the derivatives of both sides of (6) coincide at $z=1$. By analytic continuation (6) holds for $z\notin(-\infty,0]$. ∎

From (6), using Stirling’s formula, the residue theorem, and changing $z$ into $1-z$, after a few simplifications one easily gets (5), very much as in the standard proofs of (3) and (4). Of course, it is possible to go the other way, which is the usual proof of (6).

Let us finish this short paper with two formulas formally close to (6): the first one (see [2, p.43]) is

[TABLE]

and is used in the standard proof of the second Barnes lemma. As to the second one, let us consider (see [3, Vol I, p.193]) the sequence of polynomials

[TABLE]

Here $\xi$, $p$ and $v$ are complex numbers and $p+n+1$ is admissible. Then

[TABLE]

where $0<{\rm Re}\,\sigma<{\rm Re}\,q$ and $0<{\rm Re}(\xi-\sigma)<{\rm Re}(q-p)$. It is worth noticing that the generating function of the sequence $H_{n}$ is

[TABLE]